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2 years ago

Which of the following is NOT common of elite Shang burials?

Physics

1 answer:

Zigmanuir [339]2 years ago

6 0

Answer:

Large above ground mausoleums were not common in the elite Shang burials.

Explanation:

Large, above the ground mausoleums were not common so the answer is option B.

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Which change indicates that the universe is expanding?

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Red shift of distant galaxies

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3 years ago

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The body contains many small currents caused by the motion of ions in the organs and cells. Measurements of the magnetic field a

RoseWind [281]

Answer:

270 μA

Explanation:

Use the magnetic field due to long, straight wire and solve for current I.

$B= \frac{\mu_0I}{2\pi r}$

$I= \frac{2\pi r B}{\mu_0}$

plug in the values

$I= \frac{2\pi(5.40\times10^{-2})\times10\times10^{-6}\times10^{-4}}{4\pi\times10^{-7}}$

= 2.7×10^{-4)×10^6

=270 μA

The current that flows in the heart is 270 μA

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3 years ago

If arts gravitational pool just got three times stronger what would happen to our weight

IRINA_888 [86]

We would just get three times heavier.

weight = mass x gravity (9.8 m/s/s)

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3 years ago

A particular cylindrical bucket has a height of 36.0 cm, and the radius of its circular cross-section is 15 cm. The bucket is em

Sergeeva-Olga [200]

Answer:

a. 0.000002 m

b. 0.00000182 m

Explanation:

36 cm = 0.36 m

15 cm = 0.15 m

a) We can start by calculating the air-water pressure of the bucket submerged 20m below the water surface:

$P = \ro g h = 1000 * 10 * 20 = 200000 Pa$

Suppose air is ideal gas, then if the temperature stays the same, the product of its pressure and volume stays the same

$P_1V_1 = P_2V_2$

Where P1 = 1.105 Pa is the atmospheric pressure, V_1 is the air volume in the bucket on the suface:

$V_1 = Ah$

As the pressure increases, the air inside the bucket shrinks. But the crossection area stays constant, so only h, the height of air, decreases:

$P_1Ah_1 = P_2Ah_2$

$h_2 = h_1\frac{P_1}{P_2} = 0.36\frac{1.105}{200000} = 0.000002 m$

b) If the temperatures changes, we can still reuse the ideal gas equation above:

$\frac{P_1Ah_1}{T_1} = \frac{P_2Ah_2}{T_2}$

$h_2 = h_1\frac{P_1T_2}{P_2T_1} = 0.36\frac{1.105 * 275}{200000*300} =0.00000182 m$

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3 years ago

You have a concave mirror of focal length 20 cm. You want to obtain a real image of double the

ExtremeBDS [4]

Between centre of curvature and principal focus.

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2 years ago